You have found the following ages (in years) of all 5 zebras at your local zoo: $ 25,\enspace 10,\enspace 4,\enspace 21,\enspace 1$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 5 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{25 + 10 + 4 + 21 + 1}{{5}} = {12.2\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $25$ years $12.8$ years $163.84$ years $^2$ $10$ years $-2.2$ years $4.84$ years $^2$ $4$ years $-8.2$ years $67.24$ years $^2$ $21$ years $8.8$ years $77.44$ years $^2$ $1$ year $-11.2$ years $125.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{163.84} + {4.84} + {67.24} + {77.44} + {125.44}} {{5}} $ $ {\sigma^2} = \dfrac{{438.8}}{{5}} = {87.76\text{ years}^2} $ The average zebra at the zoo is 12.2 years old. The population variance is 87.76 years $^2$.